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Characterisation of Inegrals

We want to know what are the sufficient and necessary conditions for Riemann and Lebesgue integrals.

Below we consider a function bounded defined on a closed bounded interval .

Lemma 1: Let and be sequences of integrable functions over s.t. for every and every , and

. Furthermore, is increasing and is decreasing. Then is integrable over and .

Proof: Let and . Since a monotone sequence converges, the above is well defined, and furthermore we have

, for all and all .
Therefore,

. Since the right-hand-side of (1) converges to 0, we have . Thus, a.e. on . Equivalently, we have that converges to pointwise a.e. on . So is measurable on . Since and are integrable over , by Comparison Test is also integrable over .

For all , we have

. Hence, we get . Similarly, we get .

Theorem 2: Let be a bounded function on a set of finite measure. Then is Lebesgue integrable, if and only if, is measurable on .

Proof: If is measurable, previsouly we have already proved that is Lebesgue integrable, by applying approximation of simple functions.

Conversely, suppose is Lebesgue integrable over . By definition of Lebesgue integrals, for every there are simple functions and s.t. and . Hence, . Next, we let and . Observer and are also simple functions, and is increasing and is decreasing. By Lemma 1, we have pointwise a.e. on . Therefore, is measurable.

Riemann Integral

Let’s review again the definition of Riemann integral. Let be a bounded function defined on a closed interval , where are real numbers. We say a sequence is a partition of , if .

Let be a parition, and let . For each , define and . So the “upper sum” with respect to is defined as

. Similarly, the “lower sum” with respect to is defined as

. Now the upper Darboux sum is defined as,

, and the lower Darboux sum is defined as

. If the two sums agree, we say is Riemann integrable, and write

It is not difficult to show that

Theorem 3: If is Riemann integrable over , then is Lebesgue integrable over and the two integrals are equal.

Proof: Skipped.

Now we prove the characterization of Riemann integrals.

Theorem 4: Let be a bounded function on a closed interval . Then is Riemann integrable, if and only if, is continuous almost everywhere on .

Proof: Suppose is Riemann integrable. For every , there are paritions such that the difference of upper and lower sums are very small, i.e.

. Note that if a partition is a refinement of another partition, its upper sum decreases and its lower sum increases. Let be the common parition of (i.e. take the union of the partition points of and order them in ascending order), and similarly let be the common partition of . Finally, let be the common parition of . So we arrive at a sequence of paritions where is a refinement of . Furthermore, we maintain the property that

. Define for and . Similarly, define for and . Then is increasing and is decreasing. are step functions and their Riemann integrals equal their Lebesgue integrals. By (1) we know

, as . Hence, by Lemma 1 we know and pointwise almost everywhere.

Let be the set where pointwise convergence fails or no parition points of exists. Note is countable. We claim is continuous on .

In fact, let . For any there is some such that

. Therefore, we choose and for any we have

. By (2) and (3), . Thus, is continuous at and we complete the first part of the theorem.

Conversely, suppose is continuous almost everywhere. Let be any sequence of paritions of such that . We claim

. If (4) is verified, then

, and the proof will be completed.

Let be the lower step function associated with and the upper step function. The Rimann integral of a step function equals its Lebesgue integra. So to verify (4) is to prove

. Note is increasing and is decreasing. So and are uniformly bounded functions. By Bounded Convergence Theorem, to prove (6) it suffices to show that and pointwise at the set of points at which is continous and which are not parition points of any partition .

Let be such a point. Since is continuous at , for any there is some s.t. when we have

. Since , there is some s.t. when we have where is an interval in and . Therefore, by (7)

. Thus, when . This completes the proof.

Fundamental Theorems of Differential Calculus w.r.t. Riemann Integrals

For Riemann integrals, besides its characterization that Rimann integrability is equivalent to continuity almost everythere, we have 2 further fundamental properties.

Theorem 5: Let be a bounded function on a closed interval . If is Riemann integrable, then is continuous at .

Proof: Note that if is Riemann integrable, is continuous a.e. on and thus continuous a.e. on where . If we take then is properly defined on .

First consider . Since is Riemann integrable, it is also Lebesgue integrable. By Lemma 10 of Note LIR 5, for any positive there is a positive s.t. whenever we have

. So take and x \in (x_0 -\frac{\delta}{2}, x_0 + \frac{\delta}{2})A = (x_0, x)x \geq x_0A = (x, x_0)x < x_0$.

Similarly, we can easily show is right continuous at and left continuous at .

Theorem 6: Let be a bounded function on a closed interval . If is continuous at then is differentiable at and .

Proof: First we show when and ,

. In the proof of Theorem 4, we have shown that there is a sequence of paritions and the associated upper functions s.t. . Therefore, for any positive there is a positive and some sufficiently large s.t. every interval in the parition satisfies and

.

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Question 1: Let be a set function defined for all sets in a -algebra with values in . Assume is countably additive over countable disjoint collections of sets in .

(a) Prove that if and are sets in with , then .

(b) Prove that if there is a set in for which , then

(c) Let be a countable collection of sets in . Prove that .

Solution: (a) Note and are disjoint. Hence . Since , we have as desired.

(b) If is the only set for which , then for any set we have . Note and are disjoint. So . It follows that .

If , then and are disjoint. So . Since , we have .

(c) If , then the result holds trivially. So we assume the right hand side is finite.

Let and when . Then . Note the sets in are mutually disjoint, and each So by assumption we have

Question 2: Prove that for any countable set , its Lebesgue outer measure .

Proof: Let be an enumeration of . Then for any , is an open cover of . Let . So by definition of Lebesgue outer measure, . Since can be arbitrarily small, we must have .

Question 3: Let be the set of irrational numbers in the interval . Prove that Lebegue outer measure of is , i.e. .

Proof: Since , . On the other hand, let be the set rational numbers. Then is countable. By Question 2, we have . So . Thus, .

Question 4: A set is called a set if it is a countable intersection of open sets. Show that for any bounded set , there is a set that .

Proof: Since is bounded, there is some such that the open set covers . Thus, . By definition of Lebesgue outer measure, for any natural number , there is an open cover and , such that . So we construct a sequence of such open sets . Now take and is a set. We claim .

To see this, observe that every . Hence and we have . On the other hand, for every . Thus for every . As a result, we must have .

Question 5: Let be the set of rationals in . Let be a finite collection of open intervals that covers . Prove that .

Proof: Without loss of generality, we may assume has these properties: (1) ; (2) they are mutually disjoint.

So if , then there is some open interval in such that is disjoint with any interval in , and contains a rational number since the set of rational numbers is dense in . This contradicts with the fact that is collection of .

Question 6: Prove that if , then .

Proof: Observe . Conversely, since , .

Question 7: Let and be bounded sets for which there is an such that for all and . Prove that .

Proof: It suffices to prove . To this end, we need to show for any collection that covers , we have .

Without loss of generality, we may assume every contains some points in or or both. Consider an interval . Let and . Note has these properties: (1) ; (2) is open; (3) no belongs to . has similar properties. Note . Since is bounded by , we may express as a finite union of open intervals . Similarly, . Hence .

Note the countable union of covers and countable union of covers . So .

Question 8: Show that if has finite measure and , then is the disjoint union of a finite number of measurable sets, each of which has measure at most .

Proof: Observe that since . Hence there is some such that . Now partition to sufficiently smaller intervals each of which has length at most .

Question 9: (Lebesgue) Let have finite outer measure. Show that is measurable if and only if for each open, bounded interval ,
.

Proof: If is measurable, by definition the conclusion is true. Conversely, assume it holds for any open, bounded interval . Since has finite outer measure, by definition of outer measure, for any there is an open set containing such that . To prove the final result, it suffices to show for this we must have

. In fact if (1) holds, then by outer approximation will be measurable.

To prove (1), note can be expressed as a countable union of disjoint open intervals, i.e. . So . By assumption, each . So

. Observe that . Hence . Hence . Similarly, . Hence . Thus, (1) holds.

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General Case

Let be a measurable function on . We write and . So . Observe that are all non-negative measurable functions on . Recall that for a non-negative measurable function , we say is integrable over if is well defined and .

Here we define the Lebesgue integration for a generic measurable function . And we state its various properties without proofs.

Lemma 1: Let be a measurable function on . Then is integrable over if and only if are integrable over . Then we define

Theorem 2 (Linearity and Monotonicity): For any , we have

And if on , then

Theorem 3 (Additivity of integral over domains):

Lemma 4 (Comparison Test): Let be measurable functions on . If a.e. on and is integrable over , then is also integrable over .

Lebesgue Dominated Convergence

Theorem 5 (Lebesgue Dominated Convergence Theorem): Let be a sequence of measurable functions on and pointwise a.e. on . Let be a measurable function that is integrable over . If dominates in the sense that on for every , then is integrable over and

Proof: It is easily shown that is measurable and a.e. on . By comparison test, is integrable over . It remains to show the validility of exchanging limit and integration.

Observe the sequence converges to pointwise a.e., and and are non-negative for all . Thus, we can apply Fatou’s Lemma and have

. By (1) we further have

. Similarly, we consider the sequence and the function , and easily show that

. By (2) and (3) we have the result as desired.

Theorem 6 (General Lebesgue Dominated Convergence Theorem): Let be a sequence of measurable functions on and pointwise a.e. on . Let be a sequence of non-negative measurable functions and pointwise a.e. on . Furthermore, assume . If for every , then is integrable over and

Proof: It is easily shown that a.e. on (otherwise, there will be some which is a contradiction). So by comparison test, is integrable over .

Next, similar to the proof in Lebesgue Dominated Convergence, we consider sequence and function . Observer for every and . By Fatou’s Lemma and some manipulation we have

. Similarly, we consider sequence and function , and obtain

.

Countable Additivity and Countinuity of Integration

Theorem 7 (Countable Additivity of Integration): Let be integrable over . Let , where are measurable and mutually disjoint. Then,

Proof: Let . Define . So pointwise and . Since is integrable over , by Lebesgue Dominated Covergence,

. By additivity of integration over domains, we have

. By (1) and (2), we have

.

Theorem 8 (Continuity of Integration): Let be integrable over .

(a) If is an increasing sequence of measurable subsets of , then

(b) If is a decreasing sequence of measurable subsets of , then

Proof: (a) Let and for . Then is a sequence of disjoint measurable subsets in . By additivity of integration in Theorem 7, we have

Observe

. So by (2) we have

. By (1) and (3) we prove part (a).

(b) Let for . Then is an increasing sequence of measurable subsets in . By the result of part (a), we have

. Observe
. Since is decreasing, . So by (4) and (5) we have

.

Vitali Convergence Theorem

Lemma 9: Let has finite measure. Then for any , is a union of finitely many disjoint measurable sets, each of which has measure at most .

Proof: Since has finite measure, . So there is a such that and . Then choose sufficient large s.t. and break to intervals and each $E_i = E \cap [-N+(i-1)\delta, -N+i\delta), i=1,2,…,M-1E_M = E \cap [-N+(M-1)*\delta, N]\blacksquare$

Lemma 10: Let be a measurable function on . If is integrable over , then for any , there is a s.t. for any and , . Conversely, let . If for any there is a s.t. for any and , then is integrable over .

Proof: Without loss of generality we may assume is non-negative. Suppose . There by definition of integration, there is a non-negative function s.t. , is bounded, has finite support in and . Since is bounded, let be such a bound. So for any , we have

So choose we have the result as desired.

Conversely, assume and there is a to meet the challenge of . By Lemma 9, where and are disjoint. So .

We say a family of measurable functions on is uniformly integrable if for any there is a such that for any and any with , .

Theorem 11 (Vitali Convergence Theorem): Let has finite measure. Let be uniformaly intergrable over . Then if pointwise a.e., then is integrable over and

.

Proof: Let meet the challenge of . Since has finite measure, by Lemma 9, where and are disjoint, s.t. for every . So for every . So by Fatou’s Lemma,

. Thus, is integrable over .

Now consider

, where and are disjoint. For any , by Egoroff’s theorem, there is a such that and whenever we have

. By uniformly integrability over , the meets the challenge of (you should know the trick by choosing some where meets the challenge of Egoroff’s theorem and meets the challenge of uniform integrability). So we have

, and

. By (1)(2)(3) we have when .

If has infinite measure, we need additional property for besides uniform integrability to justify the passage of limit under integral sign.

Lemma 12: Let be integrable over . Then for any there is a set of finite measure such that

.

Proof: Since , by definition of the integral there is a non-negative function which is bounded and has finite support in s.t. . Let the finite support be and in . Then,

.

We say the family of measurable functions tight over if for any there is a subset of finite measure s.t. for every .

Theorem 13 (General Vitali Convergence Theorem): Let be uniformaly intergrable and tight over . Then if pointwise a.e., then is integrable over and

.

Proof: For any , since is tight over there is a subset of finite measure s.t. for every . By Fatou’s Lemma, . So is integrable over .

Note has finite measure. By Vitali Convergence Theorem, is integrable over . Thus is integrable over . Furthermore, there is some s.t. whenever ,

. So

, whenever .

Before we move on, we prove two corollaries of Vitali Convergence Theorems.

Corollary 14: Let be a sequence of non-negative functions that converge to pointwise a.e. on , where . Suppose is integrable over for every . Then, , if and only if, is uniformly integrable over .

Proof: By Vitali Convergence Theorem, if is uniformly integrable over , then . Conversely, suppose . Then for any there is some s.t. when , . So for any , for . Observe that the finite set of functions is uniformly integrable over with a that meets the challenge. So choose , then for every when . Hence is uniformly integrable over .

Corollary 15: Let be a sequence of non-negative functions that converge to pointwise a.e. on . Suppose is integrable over for every . Then, , if and only if, is uniformly integrable and tight over .

Proof: Note it is possible that . By General Vitali Convergence theorem, if is uniformly integrable and tight over , then .

Conversely, suppose . For any , there is some s.t. when . Since every is integrable over , by Lemma 12 there is some of finite measure s.t. for . Thus, take . So , and is true for every . Hence is tight over .

To prove uniformly integrability over , observe that for any the above statement shows that there is a subset of finite measure s.t. for every . Since has finite measure, by Corollary 14 is uniformly integrable over . So there is a s.t. for any with , . Now consider any general subset with . If , then . If , then . If both and are non-empty, then .

Converge in Measure

We say that a sequence of measurable functions defined on converges to in measure, if for any ,

. That is, the set of “bad” points that do not converge is becoming less and less as grows. Note that this “bad” set can keep changing. Therefore, convergence in measure does not imply convergence pointwise. The converse is true when has finite measure.

Theorem 16: Let has finite measure. Let converges to pointwise a.e. on . Then converges to in measure.

Proof: By Egoroff’s theorem, for any there is a measurable subset s.t. and converges to uniformly. So for any , there is some s.t. when when for . Therefore,

, when . So when we fix , for any there is some that meets the challenge. Hence converges in measure.

Though convergence in measure does not imply convergence pointwise, but there is a sub-sequence convergence in pointwise.

Theorem 17: Let converges to in measure on . Then there is a sub sequence that converges to pointwise a.e.

Proof: Observe that there is a strictly increasing sequence of natural numbers s.t.

, when . Let . Then . By Borel-Centalli Lemma, almost all belongs to finitely many . It means almost all satisfies that there is some s.t. when , . This is equivalent to the statement that for almost any , the sequence converges to . .

Corollary 18: Let be a sequence of non-negative functions on . Suppose is integrable over for every . Then, , if and only if, converges to in measure, is uniformly integrable and tight over .

Proof: Suppose , by Corollary 15 is uniformly integrable and tight over . Note here we don’t need any condition of convergence of to justify uniform integrability and tightness. To further prove converges to in measure, we apply Chebyshev’s inequality as

for a . Since , for any there is some s.t. when , . Hence by (1)

. So , and in measure.

Conversely, suppose converges to in measure, is uniformly integrable and tight over . Assume the conclusion does not hold. Then for any there is a sub-sequence s.t. . Note also converges to in measure. By Theorem 17, there is a further sub-sequence that converges to pointwise a.e. By General Vitalli Convergence, , contradicting the assumption that .

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Overview

We want to study the notion of integration for a generic measurable function on . Here is general in terms of

  • takes value in , but can be unbounded.
  • The domain , on which is defined, is measurable but can have infinite measure

As in Royden’s book, we will establish the integration of a general case step by step:

  • Define Lebesgue integration for simple functions on any measurable set
  • Define Lebesgue integration for bounded functions on with finite measure
  • Define Lebesgue integration for non-negative functions on with any measure (can be infinite)
  • Define Lebesuge integration for general cases

For each scenario, we will:

  • establish linearity and monoticity
  • establish additivity of integral over domains
  • study convergence: under what conditions we can exchange and , i.e. if in some sense, when can we have

Also note that in general cases, the integral can be . Hence we are interested in cases where . We would call this case as “ is integrable over E“.

Lebesgue Integration for simple functions

In this note, we may skip the proofs of some simple lemmas and theorems.

Let be a simple functions on with finite measure, and express as

, where takes the value in the measurable subset .

Then the Lebesgue integral of over is defined as

Theorem 1 (Linearity): Let be simple functions on with finite measure. For any ,

Proof: It suffices to prove (a) , and (b) .

For (a), notice that . Therefore,

. This proves (a).

For (b), assume and . Let and . Then let . It is easily shown that only takes values in . For , define . It is easily shown that is measurable, and for any , and are disjoint. Thus, can be rewritten as

, and

. Hence

Theorem 2 (Monotonicity): Let be simple functions on . If on , then .

Proof: Since , . By linearity in Theorem 1,

.

is bounded and has finite measure

Let be a bounded function defined on a measurable set with finite measure. is bounded, i.e. there is some s.t. on .

We define its upper Lebesgue integral over as

We will often abbreviate the above to

. We defined its lower Lebesgue integral over as

Similary, we abbreviate the above to

You can easilly prove that

Lemma 3:

We say the Lebesgue integral of exists if

. We write its integral as and have

Theorem 4: Let be a bounded function over a measurable set with finite measure. If is measurable, then is integrable over , i.e. exists.

Proof: By Simple Approximation Lemma (Lemma 3 in Note LIR3), for any there exist simple functions s.t. for all , and

. Hence,

. The above inequality holds for arbitrarily large . Therefore, we must have

.

Theorem 5 (Linearity): Let be bounded measurable functions on with a finite measure. For any ,

.

Proof: Since are bounded and measurable, is also bounded and measurable (Theorem 7 in Note LIR2). Hence is integrable over by Theorem 4.

First we prove . If , observe that

, and

. Since , we have

, which leads to

.

If , it suffices to prove . By similar arguements, we can see that

. The left hand side of (1) equals to

. By (1) and (2) we obtain . This completes the proof of for any . (The case is trivial.)

Next, we proceed to prove . Observe, for arbitrary simple functions and ,

. By arbitrariness of , we must have

. Similarly we have

. By (3) and (4) we establish .

Corollary 6 (Monotonicity): Let be bounded measurable functions on with a finite measure. If on all of , then .

Proof: Observe . Then is a non-negative measurable function on . Then where on . Thus .

Corollary 7: If is bounded and integrable over a finite measure , then

Proof: is bounded and integrable over . Observe . So by linearity and monotonicity, we have
$$

  • \int_E |f| \leq \int_E f \leq \int_E |f|
    $$
    , which leads to the desired result.

Corollary 8 (Additivity of Integral over Domains): Let be a bounded and measurable function on a set of finite measure . Let be a measurable subset. Then

Proof: Define function as on and on . Similarly, define on and on . Then it is easily shown that are both measurable on . Observe , and

. So by linearity we have

.

Corollary 10: Let be a bounded and measurable function on a set of finite measure . If , then .

Proof: For any , its integral .

Theorem 9 (Uniformly Convergent Sequence): Let be a sequence of bounded and measurable functions on a set of finite measure . If uniformly, then

Proof: For any , since uniformly, there exists some s.t. . So

.

If pointwise a.e. on , then we need some extra conditions to bound .

Theorem 10: (Pointwise Convergent and Uniformly Boundeded Sequence): Let be a sequence of bounded and measurable functions on a set of finite measure . If pointwise a.e. on and in addition is uniformly bounded, i.e. there is some s.t. on for all , then

Proof:

First by Theorem 1 in Note LIR 3, is measurable on and thus integrable over .

Let be the set that does not converge to , and . Note . If we were able to establish the result for , then

, since by Corollary 10, and are all .

Hence we may assume pointwise on . It is easily shown that . So By Egoroff’s Theorem, for any , there is a and a closed set s.t. and whenever . Therefore,

.

is non-negative and E has infinite measure

For a measurable function defined on a measurable set , we say has a finite support ,
if for , for all , and . We say
vanishes outside .

If has a finite support of and is bounded in , then we define .
The latter is well defined in the last section.

To define when and can have infinite measure, we use a set of non-negative bounded
functions that have finite support to “approach” it. Formally,

When , we say is integrable over .

Theorem 11 (Equivalent Definition): If is a non-negative measurable function defined on a measurable set
, then

Proof: Let ,
and let .
Clearly , and thus . To prove the converse, we need to show for
every , there exists some s.t. .

Note by Simple Approximation Theorem, there is a sequence in s.t.
pointwise a.e. on and for all . Without loss of generality, we have
pointwise on , where is the finite support of .

If is bounded on , then is uniformly bounded on .
So

If is unbounded on , then let and .
So for .

Since is bounded, is uniformly bounded on . Hence

On the other hand, is bounded on hence .
So there is some such that
and

Note is increasing and . By (1) and (2) we can choose sufficiently large such that
.

Theorem 12 (Lineraity and Monotonicity): Let be non-negative measurable functions over a measurable
set . For any ,

If for all , then

Proof: First let’s prove . For short, let’s write
and .
We want to show . By linearity of integral of
bounded functions, for any we take . Then . So we can easily show
.

Conversely, for any s.t. , we take . Then .
Thus . So we have established
.

Next we prove . Consider any with finite support ,
and any with finite support . Take . Then vanishes outside and
we can easily show that . Clearly
on . Hence we have

. Conversely, consider a with finite support . Let s.t.
on . Let . So on . Now take and on and on . So on and on . Also on . Thus, . This proves
.

By (1) and (2) we prove the linearity of integration.

Finally, assume . Then any would satisfy . By definition of the integral, the supremum of left-hand-side must be less than or equal to the supremum of the right-hand-side. Hence .

Corollary 13 (Additivity of Integral over Domains): Let be a non-negative measurable function on . Let be a measurable subset in . Then

Proof: Define on and on . And define on and on . Then on . By linearity, we have . Note the finite support of is . So it is not difficult to show . Similarily we have .

Theorem 14 (Fatou’s Lemma): Let be a sequence of measurable functions that converges to on pointwise a.e. Then is measurable on and

Proof: Without loss of generality, we may assume pointwise. We need to show for any bounded non-negative mesurable functions with finite support, .

To this end, let be the finite support of and let on . It is easil shown that that pointwise. Since is bounded, is bounded for any . Hence by convergence theorem, we have

. On the other hand, . By the definition of , we have . Therefore,

.

Theorem 15 (Monotone Convergence Theorem): et be an increasing sequence of measurable functions that converges to on pointwise a.e. Then

Proof: By Fatou’s Lemma, we get . It reamins to show

. Since is increasing, . So . Thus we have (1) as desired.

Theorem 16 (Chebyshev’s Inequality) For any ,

Proof: Let . We only prove the case when . Let . Then is a bounded non-negative functon with finite support. Clearly on . So

.

Corollary 17: if and only if a.e. on

Proof: If a.e. on , then any bounded non-negative with finite support must satisfy a.e. on . Then and thus .

Conversely, let . By Chebyshev’s inequality, for every natural number . Note . So , i.e. a.e. on .